| Expected Value
- Discrete Case: \mathsf {E(x)= \sum x_{i} \cdot P(X=x_{i})}
- Continuous Case: \mathsf {E(x)= \int_{-\infty}^{\infty} x \cdot f(x)dx }
- Properties (a, b \in \mathbb{R})
If X\geq 0, then E(X)\geq 0
Proof
\mathsf { E(X)=\sum_{x}x\cdot P(X=x)=\sum_{x>0} x\cdot P(X=x)\geq \sum_{x>0} 0 \cdot P(X=x)=0}
E(aX) = a E(X)
Proof
\mathsf {E(aX)=\sum _{x}a\cdot x \cdot P(X=x)=a \sum_{x} x \cdot P(X=x)=a \cdot E(x) }
E(X+Y) = E(X) + E(Y)
| Variance
- \mathsf{ Var= E(X^2)- E(X)^2 }
Proof
\mathsf{Var= E[x- E(x)^2] = E[(x- \mu)^2)]= \sum_{x} (x_{i}-\mu)^2 \cdot P(X=x)}
\mathsf{= \sum_{x}(x^2 - 2\mu x + \mu^2) \cdot P(X=x)}
\mathsf{ = \sum_{x}x^2\cdot P(X=x)-2\mu \cdot \sum_{x}x\cdot P(X=x)+ \mu^2 \sum_{x}P(X=x)}
\mathsf{=E(x^2)-2\mu^2+\mu^2 = E(x^2)-\mu^2 = E(x^2)-E(x)^2}
- Properties (a, b \in \mathbb{R})
Var(a)=0 (All values are same, then there is no variance)
Var (aX+b)= a^2 \cdotVar(x)
Proof
From \mathsf {E(aX+b)=aE(X)+b}, and \mathsf {E[(aX+b)^2]=a^2E(X^2)+2abE(X)+b^2}
\mathsf {Var(aX+b)=E[(aX+b)^2]-[E(aX+b)]^2}
\mathsf {=a^2E(X^2)+2abE(X)+b^2-[aE(X)+b]^2}
\mathsf {=a^2E(X^2)+2abE(X)+b^2-[a^2(E(X))^2+2abE(X)+b^2]}
\mathsf {=a^2[E(X^2)-E(X)^2]=a^2\cdot Var(X)}
Var(X+Y)= Var(X)+Var(Y)+ 2Cov(X,Y)
Proof
From \mathsf {E(X+Y)=E(X)+E(Y)}, and \mathsf {E[(X+Y)^2]=E(X^2)+2E(XY)+E(Y^2)}
\mathsf {Var(X+Y)=E[(X+Y)^2]-[E(X+Y)]^2}
\mathsf {=E(X^2)+E(Y^2)+2E(XY)-[ (E(X))^2 + (E(Y))^2 +2E(X)E(Y)] }
\mathsf {=E(X^2)-(E(X))^2+E(Y^2)-(E(Y))^2+2 [E(XY)-E(X)E(Y)] }
\mathsf {=Var(X)+Var(Y)+2Cov(X,Y)}
Var(X+Y)= Var(X)+Var(Y), iff X and Y are uncorrelated
| Covariance
- A measure of how much two random variables change together!
- Cov(X,Y)=E{ [X-E(X)][Y-E(Y)] } = E(XY)-E(X)E(Y)
|If X and Y are independent, then
- P(X=x, Y=y)=P(X=x)P(Y=y)
- E(XY)=E(X)E(Y),
\mathsf {E(XY)=\sum_{x,y}xy \cdot P(X=x,Y=y)= \sum_{x}\sum_{y}xy \cdot P(X=x)P(Y=y)}
\mathsf {=\sum_{x} x\cdot P(X=x) \cdot \sum_{y}y \cdot P(Y=y)=E(X)E(Y)}
- Cov (X,Y)=0
Proof
E(XY)-E(X)E(Y)=E(X)E(Y)-E(X)E(Y)=0
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